Question: In the country of United States of Heightlandia, the height measurements of ten-year-old children are approximately normally distributed with a mean of $54.3$ inches, and standard deviation of $3.1$ inches.
What is the probability that the height of a randomly chosen child is between $55.75$ and $61.25$ inches? Do not round until you get your final answer, and then round to 3 decimal places.
Answer=____________________ (Round your answer to 3 decimal places.)
1. **State the problem:** We want to find the probability that a child's height $X$ is between $55.75$ and $61.25$ inches, given $X \sim N(\mu=54.3, \sigma=3.1)$.
2. **Formula and rules:** For a normal distribution, the probability that $X$ lies between two values $a$ and $b$ is given by:
$$P(a \leq X \leq b) = P\left(\frac{a-\mu}{\sigma} \leq Z \leq \frac{b-\mu}{\sigma}\right)$$
where $Z$ is the standard normal variable with mean $0$ and standard deviation $1$.
3. **Calculate the Z-scores:**
$$Z_a = \frac{55.75 - 54.3}{3.1} = \frac{1.45}{3.1} = 0.4677419355$$
$$Z_b = \frac{61.25 - 54.3}{3.1} = \frac{6.95}{3.1} = 2.2419354839$$
4. **Find the probabilities from the standard normal distribution table or using a calculator:**
$$P(Z \leq 0.4677) \approx 0.6808$$
$$P(Z \leq 2.2419) \approx 0.9875$$
5. **Calculate the probability between the two Z-scores:**
$$P(0.4677 \leq Z \leq 2.2419) = P(Z \leq 2.2419) - P(Z \leq 0.4677) = 0.9875 - 0.6808 = 0.3067$$
6. **Final answer:** The probability that a randomly chosen child has a height between $55.75$ and $61.25$ inches is approximately **0.307** (rounded to 3 decimal places).